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# Solve the given integral $\int{\dfrac{1+{{x}^{4}}}{{{\left( 1-{{x}^{4}} \right)}^{\dfrac{3}{2}}}}\,dx}$A. $\dfrac{x}{\sqrt{1-{{x}^{4}}}}+c$B. $\dfrac{-x}{\sqrt{1-{{x}^{4}}}}+c$C. $\dfrac{2x}{\sqrt{1-{{x}^{4}}}}+c$D. $\dfrac{x}{\sqrt{1-{{x}^{4}}}}+c$

Last updated date: 15th Jul 2024
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Hint: Here we have to integrate the given function with respect to x. There is no direct formula to integrate the given function. So first we try to simplify, then if it is possible to solve it by substitution method, we substitute the quantities.

Complete step by step solution:
Finding the area of the curve's undersurface is the process of integration. To do this, cover the area with as many little rectangles as possible, then add up their areas. The total gets closer to a limit that corresponds to the area under a function's curve. Finding an antiderivative of a function is the process of integration.
Now consider the given function
$\int{\dfrac{1+{{x}^{4}}}{{{\left( 1-{{x}^{4}} \right)}^{\dfrac{3}{2}}}}\,dx}$
Take ${{x}^{2}}$ as common in the denominator
$\int{\dfrac{1+{{x}^{4}}}{{{\left( {{x}^{2}}\left( \dfrac{1}{{{x}^{2}}}-{{x}^{2}} \right) \right)}^{\dfrac{3}{2}}}}\,dx}$
The power is applicable for both terms
$\int{\dfrac{1+{{x}^{4}}}{{{x}^{3}}{{\left( \dfrac{1}{{{x}^{2}}}-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}\,dx}$
Take ${{x}^{3}}$ as common in the numerator
$\int{\dfrac{{{x}^{3}}\left( \dfrac{1}{{{x}^{3}}}+x \right)}{{{x}^{3}}{{\left( \dfrac{1}{{{x}^{2}}}-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}\,dx}$
Now cancel ${{x}^{3}}$ which is present both numerator and denominator.
$\int{\dfrac{\left( \dfrac{1}{{{x}^{3}}}+x \right)}{{{\left( \dfrac{1}{{{x}^{2}}}-{{x}^{2}} \right)}^{\dfrac{3}{2}}}}\,dx}$
Now we will substitute the denominator value as t.
$\left( \dfrac{1}{{{x}^{2}}}-{{x}^{2}} \right)=t$
On differentiating the above function we have
$\left( \dfrac{-2}{{{x}^{3}}}-2x \right)dx=dt$
Take -2 as common so we have
$-2\left( \dfrac{1}{{{x}^{3}}}+x \right)dx=dt$
So if we substitute the denominator we get the function which is present in the numerator. $\int{\dfrac{-1}{2{{\left( t \right)}^{\dfrac{3}{2}}}}dt\,}$
The constant can be written outside the integral sign
$\dfrac{-1}{2}\int{\dfrac{1}{{{\left( t \right)}^{\dfrac{3}{2}}}}dt\,}$
We shift the t power of 3 by 2 to the numerator
$\dfrac{-1}{2}\int{{{\left( t \right)}^{-\,\dfrac{3}{2}}}dt\,}$
As we know that $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c}$, by using this formula the integration of the given function will be
$\dfrac{-1}{2}\left[ \dfrac{{{t}^{-\dfrac{3}{2}+1}}}{\dfrac{-3}{2}+1} \right]$
On simplification we get
${{t}^{\dfrac{-1}{2}}}+c$
$\dfrac{1}{{{t}^{\dfrac{1}{2}}}}+c$
Now resubstituting the value of t we have
$\dfrac{1}{{{\left( \dfrac{1}{{{x}^{2}}}-{{x}^{2}} \right)}^{\dfrac{1}{2}}}}+c$
On simplifying we get
$\dfrac{x}{\sqrt{1-{{x}^{4}}}}+c$
Hence option A is the correct one.

Note: During the substitution we have converted the given function which will be in terms of x to other variables such as t or u, etc,. We can solve the integration by some standard formulas of integration directly. Otherwise we have to see alternate methods to solve it like substitution, or by taking partial fractions and so on.